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A cubical block of glass of refractive i...

A cubical block of glass of refractive index `n_1` is in contact with the surface of water of refractive index `n_2.` A beam of light is incident on vertical face of the block. After refraction a total internal reflection at the base and refraction at the opposite face take place. The ray emerges at angle `theta` as shown. The value of `theta` is given by

A

`"sin"thetaltsqrt(n_(1)^(2)- n_(2)^(2))`

B

`"tan"theta lt sqrt(n_(1)^(2) - n_(2)^(2))`

C

`"sin"theta lt (1)/(sqrt(n_(1)^(2) - n_(2)^(2)))`

D

`"tan"theta lt (1)/(sqrt(n_(1)^(2) - n_(2)^(2)))`

Text Solution

Verified by Experts

The correct Answer is:
A

`n_(1)"sin"theta_(C) = n_(2) implies "sin"theta_(C) = (n_(2))/(n_(1))`

`1 xx "sin"theta= n_(1)"sin"(90-theta_(C))= n_(1)"cos"theta_(C) = n_(1)sqrt(1-(n_(2)^(2))/(n_(1)^(2)))`
`thetasin^(-1)(sqrt(n_(1)^(2)-n_(2)^(2))), "sin" theta lt sqrt(n_(1)^(2)- n_(2)^(2))`
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